Mathematics - 2020-05-25 (page 2 of 2)

What is your mathematical status, as it were?

From the pig snout it's hard to tell.

uh, i work in industry now, did grad school in math

Aha.

i guess my general mathematical maturity would be around a generic grad student now

though somewhat rusty

but it's still something interesting that i want to keep/extend

9:27 PM

we should create a website like math stack exchange

i mean, ideally you can find a group of people at similar level with similar interests

the way stackexchange/stackoverflow is now, makes that very hard

just finding questions is not too easy tbh

I can do it

Hi, can someone help me on sum of Darboux and Riemann integrability ?

if $f:[a,b]\to\mathbb{R}$ and $P$ is an uniforme subdivision of [a,b]

why

$\int_a^b f(x)dx=\lim_{n\to+\infty} S_n(f)=\lim_{n\to+\infty} s_n(f)$

where $S_n=\frac{b-a}{n}\sum_{i=1}^n M_i$ and $s_n=\frac{b-a}{n}\sum_{i=1}^n m_i$

What is the hypothesis on $f$?

$M_i=\sup\limits_{x\in[x_{i-1},x_i]}f(x)$ and $m_i=\inf\limits_{x\in[x_{i-1},x_i]}f(x)$

9:37 PM

I know what those are.

@TedShifrin f is Riemann integrable

And what's your definition of Riemann integrable?

$s(f,P)-S(f,P)<\varepsilon$

for all $\varepsilon>0$

With some words to go along with it. That's sloppy.

For every $\varepsilon$, there exists a partition $P$.

So what is $S_n-s_n$ for your uniform partition?

@pig I definitely agree with you. I'm going to work on this :)

9:40 PM

$\frac{b-a}{n}\sum_{i=1}^n (M_i-m_i)$ @TedShifrin

And what can you do with that?

@ted was working on more elliptic curves stuff and coming to grips with how little i actually knew

I don't know

Well, I no longer know anything.

Grüß Gott

9:43 PM

@TedShifrin you can't give me a hint ?

for instance: it took me a bit to wrap my brain around $x$ having a double zero / double pole on the elliptic curve $y^2=x^3-x$ (to take a specific example)

i get it now but that was a big wtf moment

@TedShifrin can I write that $M_i=f(\alpha_i)$ ?

and $m_i=f(\beta_i)$

@linda I guess the best thing to do is to use the definition you have for Riemann integrability. Given $\varepsilon$, there is a partition $P_0$. Now relate $S_n-s_n$ to $S(P)-s(P)$ a concrete way.

I don't understand you

random question not related to math: can someone speak in a language to an italian, and the italian thinks it sounds like italian but it's actually not any language it just sounds like italian to the italian

9:47 PM

Draw a picture of your partition $P_0$ and the uniform partition $P_n$. What happens when $n\to\infty$?

I could understand a non-italian thinking it sounds like italian

@geocalc33 that would be cool if you make something like that :P

@pig yeah :)

it does remind me a little of the old chinese room idea

chines eroom?

9:52 PM

Chinese room

The Chinese room argument holds that a digital computer executing a program cannot be shown to have a "mind", "understanding" or "consciousness", regardless of how intelligently or human-like the program may make the computer behave. The argument was first presented by philosopher John Searle in his paper, "Minds, Brains, and Programs", published in Behavioral and Brain Sciences in 1980. It has been widely discussed in the years since. The centerpiece of the argument is a thought experiment known as the Chinese room.The argument is directed against the philosophical positions of functionalism...

oh ic

grabbing a description from another site:

"Searle imagines himself alone in a room following a computer program for responding to Chinese characters slipped under the door. Searle understands nothing of Chinese, and yet, by following the program for manipulating symbols and numerals just as a computer does, he sends appropriate strings of Chinese characters back out under the door, and this leads those outside to mistakenly suppose there is a Chinese speaker in the room."

@TedShifrin P_n is more large then P_0

But eventually a lot of the intervals in $P_n$ lie inside intervals of $P_0$, so that part of the contribution to $S_n-s_n$ is controlled by $S_0-s_0$. What causes problem?

what it shares with your proposal is the notion of one person perceives that they're conversing in a certain language, when in fact they're the only one who actually understands what they're saying

9:58 PM

that's interesting

10:08 PM

@TedShifrin I don't see it like this

Well, I'm telling you a way to do the proof.

ok thank you very much

10:41 PM

@Thorgott ich hab eine Frage zur deutschen Sprache! Und zwar, wenn man sagt, dass z.B. eine Funktion $f$ durch etwas festgelegt ist, heißt das dasselbe wie "$f$ ist durch blah bestimmt"?

ich vermute schon, also geh ich davon aus hahaha

ja, das ist damit gemeint

z.B. "eine lineare Funktion ist durch die Werte auf einer Basis festgelegt"

Okay good hahaha

@TedShifrin I found that the inverse implication is right when f is continuous what it means?

ah no wait I'm being a fool

Are you asking what inverse implication means?

10:54 PM

someone end my suffering

loool

Better call @Balarka.

I do not want a sheaf-theoretic interpretation

I did not just enter the chat

Mike Miller

awful notation

what are you doing

10:56 PM

mutative algebra

if f is continuous and the limits are equal then f is riemann integrable? @TedShifrin

Mike Miller

ah Module not Manifold

yeah, it's restriction/extension of scalars

I don't see the point, @linda. If $f$ is continuous, it's automatically integrable on a closed interval.

@TedShifrin perhaps it means if f is continuous then the limits are equal

10:59 PM

Well, yes, because it's integrable.

see: les-mathematiques.net/a/d/a/…

see remark before somme de riemann

Notice that their definition of Riemann integrable is not what you told me. But I don't see what the point is. If $f$ is continuous, as I said, it's Riemann integrable no matter what the definition, so everything follows.

11:32 PM

@Thorgott oh nice

six functors strikes back

also you probably meant mutative coalgebra

lol

fair

Mike Miller

I'm sorry to inform you that they are cocommutative

F lower shriek

to pay respects

nah, then they would be mmutative

not disassconcommutative?

11:41 PM

not inter universal?

intergalactic

yo what are the six functors in case of modules? if $f : A \to B$ is a ring homomorphism, $M$ is an $A$-module, $f_* M$ is just $M \otimes_A B$ and if $N$ is a $B$-module, $f^* N$ is just restriction of scalars. What is $f_! M$ man?

other way round

or is this some convention thing

Ah no you're right

It's a sheaf over Spec of the ring

And Spec is contravariant

I think $f_!$ is not necessarily quasi-coherent in general, so it just isn't a module

if $f$ is finite, then the map on schemes is proper, so in that case $f_!$ just agrees with $f_*$

boring